$\frac{yz-x^2}{1-x} = \frac{xz-y^2}{1-y}$ $\frac{yz-x^2}{xz-y^2} = \frac{1-x-1+y}{1-y}$ $\frac{yz-x^2-xz+y^2}{xz-y^2} = \frac{y-x}{1-y}$ $\frac{(x+y+z)(y-x)}{xz-y^2} = \frac{y-x}{1-y}$ $\frac{xz-y^2}{1-y} = x+y+z = \frac{yz-x^2}{1-x}$

Just this: $\frac{yz-x^2}{1-x}=\frac{zx-y^2}{1-y}=\frac{(yz-x^2)-(xz-y^2)}{(1-x)-(1-y)}=x+y+z(x\neq y)$

Let $\frac{yz-x^2}{1-x}=\frac{xz-y^2}{1-y}=d$. This gives us the next system of equations: \begin{align*} yz-x^2&=d(1-x)\\ xz-y^2&=d(1-y) \end{align*}Therefore we have $d(x-y)=d(x-1)+d(1-y)=x^2-y^2+xz-yz=(x-y)(x+y+z)$, hence $(x-y)(x+y+z-d)=0$. But given that $x\neq y$, we deduce $x+y+z-d=0$, so, we conclude $d=x+y+z$, as desired.